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In summary: The heavier mass M will always fall faster than the lighter mass m because it experiences a higher gravitational force.https://www.math.niu.edu/~cameron/courses/physics/gravitation/falling-objects.pdf

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https://en.m.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

Often when we try to think our way through these things without fully understanding the math, we get it wrong and that’s what has happened here.

Here’s a simple discussion on Newton’s law

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jedishrfu said:

https://en.m.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

Often when we try to think our way through these things without fully understanding the math, we get it wrong and that’s what has happened here.

Here’s a simple discussion on Newton’s law

Okay, this is all true, but consider the following scenario.

You have 2 planets will identical mass and no other forces are acting on the objects is the gravitational force between them. We will name these planets P1 and P2. Because the objects have the same mass as Earth, the objects accelerate to then will be 9.81 m/s^2. So P2 will be falling towards P1 9.81m/s^2. But P1 will be falling towards P2 at the same rate as well. The net rate that the the two planets accelerate towards each other will be 19.62m/s^2 because each object has the same mass and each object is expirencing the same force. Because the objects we experiment with are so small relative to the mass of Earth, the net acceleration towards one another is extremely close to 9.81m/s^2, but the large the mass gets relative to the mass of earth, the higher the net acceleration will be.

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Just a random thought I had and wanted feedback on! Thank you!Bandersnatch said:

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I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?Ty Wendland said:Just a random thought I had and wanted feedback on! Thank you!

They do, the experiment has been done and measured countless times

Astronauts on one of the moon landings even did it on the moon

The classic hammer and feather experiment

On Earth

The experiment was done in a vacuum chamber to negate air resistance and both the hammer and feather hit the ground together

Prof. Brian Cox did an experiment at a NASA vacuum chamber with a bunch of feathers and a bowling ball

https://www.bing.com/videos/search?...7E3D58E6A738E5D18D5B7E3D58E6A738E5D&FORM=VIREDave

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Me too.davenn said:I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?

There is a great temptation to look at these things from the point of view that says "Science got it wrong". I feel that is unhealthy because it neglects the basic authority of what is, in modern times, a very well justified body of work. In all matters Scientific, the actual numbers count and you have to take account of that before thinking that someone has found a loophole. Galilieo's (thought?) experiment shows that 'things fall down with the same acceleration'. That has been demonstrated to be true to greater and greater accuracy over the years.

What happens between bodies of near equal masses (planets etc.), follows the

The attractive force between two masses m and M is always F = mMG/d

So why is it that that Earth seems to be different from the falling ball? Simply because the two accelerations are

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Physics being a quantitative science, it's worth calculating the magnitude of this effect and comparing it with all the other experimental uncertainties in any measurement of the time it takes a dropped object to hit the ground. (For example... suppose someone on the other side of the Earth happens to fire off an antiaircraft gun while you're running your experiment? And just where is Jupiter right now?)Ty Wendland said:Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

As for whether all of this means that it is wrong to say that things "fall at the same speed regardless of their weight"? It's hard to improve on Isaac Asimov's take on this sort of question: https://chem.tufts.edu/answersinscience/relativityofwrong.htm

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The same is true for two small object falling on exactly opposite sides of the Earth.Ty Wendland said:The net rate that the the two planets accelerate towards each other will be 19.62m/s^2.

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Ty Wendland said:Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

If I remember correctly: Relative to earth, a heavier object would in principle accelerate faster and thus hit the ground after a shorter time compared to a lighter one. This doesn't hold, however, if the the body originates from the Earth itself. All terrestrial bodies falling individually fall at exactly the same rate in an ideal scenario.

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As worded the logic of the OP looks fine (with the possible exception of the last sentence), it just doesn't include numbers. I would hope and assume the OP recognizes that a hammer isn't going to accelerate the Earth very much.davenn said:I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?

They do, the experiment has been done and measured countless times

Also, there is a trick to this question depending on the wording in that two objects simultaneously dropped to Earth will impact at the same time, but if dropped separately the fall times will be (very, very slightly) different.

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sophiecentaur said:Me too.

There is a great temptation to look at these things from the point of view that says "Science got it wrong". I feel that is unhealthy because it neglects the basic authority of what is, in modern times, a very well justified body of work. In all matters Scientific, the actual numbers count and you have to take account of that before thinking that someone has found a loophole. Galilieo's (thought?) experiment shows that 'things fall down with the same acceleration'. That has been demonstrated to be true to greater and greater accuracy over the years.

What happens between bodies of near equal masses (planets etc.), follows theexact same rulesand thesame basic equationsuntil General Relativity comes into it. You just have to look at the real (quantitative) consequences of that.

The attractive force between two masses m and M is always F = mMG/d^{2}. (G is the universal gravitational constant andnot g,d is the distance between the centres of mass of the two objects and both objects are actually heading for theirmutual centre of mass)The resulting acceleration of the two bodies will be F/m and F/M. Newton tells us that. Clearly the larger mass will accelerate less than the smaller mass. F/m is the same for any value of m and itonlydepends on the 'other' mass M.

So why is it that that Earth seems to be different from the falling ball? Simply because the two accelerations aresodifferent and the Earth's surface accelerates so little, relative to the centre of mass of the two. One easily measurable effect of what happens between two no-dissimilar masses is the wobble of the Earth as the Moon orbits it. The Centre of mass of the two (the Barycentre) is about 4500km from the Earth's centre, which shows that the Earthispulled into 'an orbit' by the gravity of the Moon. There are no convenient stationary large masses up there to observe how they fall together so we have to imply things from orbiting objects but the centripetal acceleration of the Earth as it orbits round the barycentre is (2Πf)^{2}r which comes to about30μm/sHow much less is the acceleration of the Earth when you drop a 1kg mass - or a 10kg - or a 1000kg? That's why we can say the Earth stays still for the Galileo experiments and why we can use g all over the place.^{2}.

I don’t think that objects accelerate to Earth at different speeds, it’s a proven fact that everything accelerated to Earth at 9.81m/s^2. What I’m focused on is the time it two objects, the Earth and some dropped object, to come into contact. Obviously the mass difference between a 1kg object and a 10kg object compared to the Earth is nearly identical, but there is still a difference. In my scenario I’m saying there are no other external forces acting on the two objects besides gravity( which I know is extremely unrealistic). It’s

more of a hypothetical thought experiment than anything I would apply to the real world.

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Ty Wendland said:I don’t think that objects accelerate to Earth at different speeds, it’s a proven fact that everything accelerated to Earth at 9.81m/s^2. What I’m focused on is the time it two objects, the Earth and some dropped object, to come into contact. Obviously the mass difference between a 1kg object and a 10kg object compared to the Earth is nearly identical, but there is still a difference. In my scenario I’m saying there are no other external forces acting on the two objects besides gravity( which I know is extremely unrealistic). It’s

more of a hypothetical thought experiment than anything I would apply to the real world.

If you do the maths then two objects of non-negligible masses ##m_1## and ##m_2## will collide in the same time that a small object of negligible mass takes to fall to a large object of mass ##m_1 + m_2##.

More generally you could search online for the two body gravitational problem and the concept of "reduced mass".

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Nugatory said:And just where is Jupiter right now?

Jupiter? From my calculations, even Pluto has a greater effect on me than I have on the Earth.

Earth's acceleration towards me: 1.23E-22 m/s^{2}

My acceleration towards Pluto: 2.59E-14 m/s^{2} (rough minimum distance)

Factor: ≈200 million

My acceleration towards Pluto: 2.59E-14 m/s

Factor: ≈200 million

Fun thought experiment.

Thank you! I've always wondered about this. Never bothered to do the maths till now.Ty Wendland said:It’s more of a hypothetical thought experiment than anything I would apply to the real world.

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OmCheeto said:Jupiter? From my calculations, even Pluto has a greater effect on me than I have on the Earth.

Earth's acceleration towards me: 1.23E-22 m/s^{2}

My acceleration towards Pluto: 2.59E-14 m/s^{2}(rough minimum distance)

Factor: ≈200 million

I guess that really shows how silly it is to calculate the amount I cause the Earth to accelerate towards me when an object billions of kilometres away has a much greater affect on me than I do the Earth!

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jkxzgyk said:

I guess I worded what I was trying to say incorrectly. You are absolutely right. I was thinking more that these “experiments” were done separately, and the times for the objects of different mass were measured separately without influencing one another. I guess this is why I posted this, is so others could find the flaws in my logic!

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If we're talking about completely negligible, but in principle present effects, then we also have the direction of the force from each of the objects being different. This means that the Earth will fall a bit sideways, away from the line equidistant from the two objects, and towards the heavier object.jkxzgyk said:

This in turn - continuing with the unmeasurable physics theme - will mean that due to the surface of the Earth curving away more for the lighter object, it will have a larger distance to cover, and will fall after the heavier one.

Just to hammer in the obvious - for the m1, m2 << M case, any such difference will be swamped by orders of magnitude-larger errors in measuring instruments, timing of the release, the fact that no floor follows exactly the equipotential surface of the planet. And pretty much anything else one can think of.

But if one imagines the masses involved as being on a sliding scale, going all the way from m1, m2 << M to m1, m2 >> M, at some point the differences in the rate of falling become important and easy to spot. After all, if we 'drop' the Mars and the Sun on the Earth from 1 AU, they won't hit the ground at the same time.

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Ty Wendland said:Okay, this is all true, but consider the following scenario.

You have 2 planets will identical mass and no other forces are acting on the objects is the gravitational force between them. We will name these planets P1 and P2. Because the objects have the same mass as Earth, the objects accelerate to then will be 9.81 m/s^2. So P2 will be falling towards P1 9.81m/s^2. But P1 will be falling towards P2 at the same rate as well. The net rate that the the two planets accelerate towards each other will be 19.62m/s^2 because each object has the same mass and each object is expirencing the same force. Because the objects we experiment with are so small relative to the mass of Earth, the net acceleration towards one another is extremely close to 9.81m/s^2, but the large the mass gets relative to the mass of earth, the higher the net acceleration will be.

You may want to start by reading this:

https://www.physicsforums.com/insights/why-is-acceleration-due-to-gravity-a-constant/

Zz.

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I pointed that out in Post #12...jkxzgyk said:

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Works for any object or assemblage ofTy Wendland said:I guess I worded what I was trying to say incorrectly. You are absolutely right. I was thinking more that these “experiments” were done separately, and the times for the objects of different mass were measured separately without influencing one another. I guess this is why I posted this, is so others could find the flaws in my logic!

Adding in an object from

Drop time is the time for the objects to reach the centre of mass.

I think you have noted that correctly in you post #13. about the two planets.

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OldYat47 said:At a human scale two objects of different masses will initially accelerate at exactly the same rate.

From a fundamental perspective regarding the effective - let's say - "closing acceleration", this would only be the case if the two objects are of terrestrial origin.

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That's what was implied by "at a human scale" although the analysis works on any planet with an atmosphere. More to come on a "non-human" scale as soon as I edit it more. Too lengthy now.Lord Jestocost said:From a fundamental perspective regarding the effective - let's say - "closing acceleration", this would only be the case if the two objects are of terrestrial origin.

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F = G*(m1+m2)/d. Since the masses are identical we could write that as:

F = (G*2*m)/d

Acceleration of each sphere:

Am1 = [(G*2*m)/d]/m1

Am2 = [(G*2*m)/d]/m2

Closing acceleration = Am1+Am2.

Aclosing = 2*(G*2m/d)/m, or A closing = (4*G/d)

Now suppose the mass of m2 is twice the mass of m1. We can substitute (2*m1) for m2.

F = G*(3m)/d

Acceleration of each sphere:

Am1 = [G*(3m)/d]/m

Am2 = [G*(3m/d]/(2*m)

Simplifying and adding those two together to get the closing acceleration:

A closing = (3*G/d) + (3/2*G/d) = 4.5*(G/d).

So for these conditions the lighter object will collide before the heavier object. Hope that's right, it's early.

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Try ##F=\frac{Gm_1m_2}{d^2}##OldYat47 said:F = G*(m1+m2)/d.

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We have two spheres of similar size and identical mass in otherwise empty space. The gravitational force attracting them to each other is:

F = G*(m1*m2)/d^2. (Using ^ as indication of exponent). Since the masses are identical we could write that as:

F = (G*m^2)/d^2

Acceleration of each sphere:

Am1 = [(G*m^2)/d^2]/m1 or Am1 = (G*m)/d^2, and since the masses are identical,

Am2 = (G*m)/d^2

Closing acceleration = Am1+Am2.

Aclosing = 2*(G*m)/d^2

Now suppose the mass of m2 is twice the mass of m1. We can substitute (2*m1) for m2.

F = G*(2m^2)/d^2

Acceleration of each sphere:

Am1 = [G*(2m^2)/d^2]/m or Am1 = 2*(G*m)/d^2

Am2 = [G*(2m^2/d^2]/(2*m) or Am2 = (G*m)/d^2

Aclosing = Am1 + Am2 = 3*(G*m)/d^2

So for these conditions the lighter object will collide before the heavier object.

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You are trying to claim that mass 1 collides with mass 2 before mass 2 collides with mass 1? That does not seem very reasonable.OldYat47 said:So for these conditions the lighter object will collide before the heavier object.

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A true comparison is when the force and distance are the same but the masses of the objects change. That is, the product of mass 1 and mass 2 is the same in both cases.

So let m1 = 2 and m2 =2.

F = (G*4)/d^2, the acceleration of each mass is (G*2)/d^2, and

A closing = (G*4)/d^2

Now let m1 = 1 and m2 = 4

F doesn't change, = (G*4)/d^2

But the accelerations do change:

Am1 = (G*4)/d^2

Am2 = (G)/d^2, and

A closing = (G*5)/d^2

So in cases where the gravitational force is equal, where masses are equal closing acceleration is lower than where masses are unequal. The greater the difference in masses, the greater the closing acceleration.

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Gravity is different and can be calculated as GMm/r^2 where G is the gravitational constant, M is one mass, m the other and r the distance between them. Because in the real world the ratio of M to m is so huge you cannot measure differences of acceleration in a gravitational field with the skin friction of an atmosphere.Ty Wendland said:

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Suppose we have two object with masses m1 and m2 = 4*m1. We have a third object with mass m3 = 10^24*m1. The objects are spherical, m1 and m2 are some distance d above the surface of m3, and there is no atmosphere.

We drop m1 onto m3.

Attractive force = [(G * (10^24 * m1) * m1)] / (d^2) or

Attractive force = (G * 10^24 * m1^2) / (d^2)

Acceleration of m1 = (G * 10^24 * m1) / (d^2)

Acceleration of m3 = (G * m1) / (d^2)

Closing acceleration (m1 to m3) = [G * m1 * (10^24 + 1)] / (d^2)

We drop m2 onto m3.

Attractive force = [(G * (10^24 * m1) * (4 * m1)] / (d^2) or

Attractive force = (G * (40^24 * m1^2) / (d^2)

Acceleration of m2 = (G * 10^24 * m1) / (d^2)

Acceleration of m3 = (G * 4 * m1) / (d^2)

Closing acceleration (m2 to m3) = [G * m1 * (10^24 + 4)] / (d^2)

So m2 and m3 would collide in less time than m1 and m3, but the difference in time is tiny (certainly beyond our capacity for measurement) until d gets quite large, and m3 moves a relatively tiny amount. This is different from my previous example because the attractive forces are different in this case.

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The algebra is not complex, but perhaps I could have included the non-simplified factors.

We have two objects, m1 and m3. m3's mass is 10^24 times m1's mass or (10^24 * m1). The attractive force is [G * m1 * m3 / d^2].

So the

The acceleration of m1 is the attractive force divided by m1 = [G * (10^24 * m1^2) / (d^2)] / m1 = G * (10^24 * m1) / (d^2) =

The acceleration of m3 is the attractive force divided by m3 = [G * (10^24 * m1^2) / (d^2)] / (10^24 * m1) =

The closing acceleration is the sum of those two accelerations,

Similar for m2 and m3.

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